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IMPORTANT 

MATHEMATICAL 

DISCOVERIES 


UC-NRLF 


$B    527    TDE 


^ 


By  P.  D.  WOODLOCK 


^  r 


IN   MEMORIAM 
FLORIAN  CAJO 


J 


R^faaetaci 


IMPORTANT  DISCOVERIES 

IN  PLANE  AND  SOLID 
GEOMETRY 


CONSISTING    OF 
THE    RELATION  OF    POLYGONS  TO    CIRCLES 

AND  THE 

EQUALIZING  OF   PERIMETERS  TO   CIRCUMFERENCES 

AND 

DRAWING  CURVED  LINES   EQUAL  TO    STRAIGHT   LINES 

THE 

TRISECTION  OF  AN   ANGLE 

AND  THE 

DUPLICATION  OF  THE  CUBE 


By  p.  D.  WOODLOCK 


nnn 


PRESS    OF 

e.  w.  stephens  publishing  company 
columbia,  missouri 

Copyright  1912 
By'-iP.  D.  Wooi/i-ooK. 


PREFACE. 

In  publishing  this  book  the  author  feels  confident  that 
he  has  added  something  to  geometrical  science,  which  has 
not  heretofore  been  known. 

He  especially  invites  geometricians  and  mathemati- 
cians to  examine  carefully  and  without  bias,  the  several 
propositions  and  problems,  and  their  demonstrations,  con- 
tained in  the  book,  and  he  has  no  doubt  that  they  w^ill 
find  interest  in  every  page. 

Some  of  the  problems  appearing  in  this  book  have  occu- 
pied the  attention  of  geometricians  in  all  ages  since  the 
introduction  of  geometry  as  a  science,  yet  all  attempts 
at  their  solution  have  been  unsuccessful  down  to  the  pres- 
ent time.  That  the  author  of  this  little  work  has  been 
rewarded  with  the  discovery  of  the  true  solutions  of  these 
problems  he  confidently  leaves  to  the  consideration  and 
candid   judgment   of   geometricians   throughout   the   world. 


iv  270590 


PART  I. 

CONSISTING    OF 

PRELIMINARY  DEMONSTRATIONS 

LEADING    TO    THE 

EQUALIZING  OF  CURVED  LINES 
TO  STRAIGHT  LINES. 


PRELIMINARY  DEMONSTRATIONS. 
PROPOSITION  A. 

To  form  a  series  of  polygons  upon  a  square  or  equilateral 
triangle  having  the  same  center  and  equal  perimeters, 
the  number  of  their  sides  being  to  each  other  consecutively 
in  the  ratio  of  two. 


Figure  A. 

In  Figure  A,  let  BCDE  be  a  square,  and  draw  the  diag- 
onals BD  and  CE,  bisecting  each  other  at  A,  which  is  the 


8 


Important  Mathematical  Discoveries 


center  of  the  square,  and  draw  the  Hne  AP.  Then  AP  is 
the  radius  of  the  inscribed,  and  AC  is  the  radius  of  the 
circumscribed  circles.  Then  bisect  the  hues  AB  and  AC 
at  R  and  L,  and  draw  the  lines  RF  and  LH  equal  to  RB 
and    LC    respectively,    and    parallel    to   AP,    and    draw  the 


Figure  B. 

lines  AW,  AX  and  Ay,  bisecting  the  lines  BE,  CD,  and 
DE,  and  draw  RN  equal  to  RB,  and  parallel  to  AW,  and 
draw  LI  equal  to  LC,  and  parallel  to  AX.  And  in  like 
manner  bisect  the  lines  AE  and  AD  at  G  and  U,  and  draw 
the  lines  CM   and   GO,   equal   to  GE,   and   draw  UK  and 


Preliminary  Demonstrations  9 

UJ,  equal  to  UD,  and  draw  the  lines  RL,  LU,  UG,  and 
GR,  and  draw  the  lines  FH,  HI,  IJ,  JK,  KO,  OM,  MN 
and  NF.  And  the  figure  thus  formed  is  an  octagon,  whose 
perimeter  is  equal  to  the  perimeter  of  the  square  BCDE. 
Now  the  line  Cr  is  perpendicular  to  LH,  and  HT  is  per- 
pendicular to  LC.     And  LH  being  equal  to  LC,  therefore 


Figure  C. 


the  line  Cr  equals  HT.  In  like  manner  FS  =  BV,  and 
FH  =  Vr.  Hence  the  lines  SF  +  FH  +  HT  =  BV  +  Vr 
+  rC  =  BC,  which  is  a  side  of  the  square.  Now  the  lines 
SF  +  FH  +  HT  are  equal  to  one-fourth  of  the  perimeter 
of   the   octagon.     And    BC   is   one-fourth    of   the   perimeter 


10 


Important  Mathematical  Discoveries 


of  the  square;  therefore  the  perimeter  of  the  octagon  FHIJ 
is  equal  to  the  perimeter  of  the  square  BCDE. 

In  Figure  B,  let  the  square  BCDE  and  the  octagon 
FHIJ  have  equal  perimeters,  and  produce  AP  to  n,  which 
bisects  FH,  and  draw  the  lines  AF  and  AH.     Bisect  AF 

c 


H 


B 


Figure  D. 

at  R,  and  AH  at  V,  and  draw  the  lines  Ra  and  Rb,  each  equal 
to  RF,  and  parallel  to  AB  and  An  respectively.  In  like 
manner  draw  the  lines  Vc  and  Vd,  each  equal  to  VH,  and 
draw  ab,  be,  and  cd,  and  draw  the  line  ae,  parallel  to  FS, 
and  dh  parallel  to  HT.  Then  ea  +  ab  +  be  +  cd  +  dh  = 
SF  +  FH  +  HT  =  BC.  And  as  ea  +  ab  +  be  +  cd  +  dh 
form  one-fourth  of  the  perimeter  of  a  polygon  of   16  sides, 


Preliminary  Demonstrations  11 

and  by  completing  the  remaining  sides  of  that  polygon, 
its  perimeter  is  equal  to  the  perimeter  of  the  octagon,  and 
also  to  that  of  the  square,  their  sides  are  to  each  other 
consecutively  in  the  ratio  of  two.  Proceed  likewise  in 
forming  polygons  of  32,  64,  128,  256,  etc.,  sides,  until  they 
become  infinite  in  number  and  minuteness. 

In  Figure  C,  let  BCD  be  an  equilateral  triangle,  and 
bisect  its  sides  at  W,  N  and  M,  and  draw  the  lines  CM, 
BN  and  DW,  intersecting  each  other  at  the  point  A,  which 
is  the  center  of  that  triangle.  And  bisect  AC  at  V,  AB 
at  T,  and  AD  at  L,  and  draw  VI  and  TH,  parallel  to  AW, 
making  VI  =  VC,  and  TH  =  TB.  And  draw  VJ  and 
LK  parallel  to  AN,  and  equal  to  VC  and  LD.  In  like 
manner  draw  LE  and  TF,  equal  to  LD  and  TB,  and  draw 
the  lines  HI,  IJ,  JK,  KE,  EF,  and  FH.  And  the  figure 
thus  formed  is  a  hexagon  whose  perimeter  is  equal  to  that 
of  the  triangle  BCD. 

Now  it  will  be  seen  that  PH  =  BS,  HI  =  SR,  and  10 
=  RC.  Hence  PH  +  HI  +  10  =  BC.  In  like  manner  OJ 
+  JK  +  KX  =  CD,  and  XE  +  EF  +  FP  =  BD.  There- 
fore their  perimeters  are  equal. 

Figure  D  represents  an  equilateral  triangle,  a  hexagon, 
and  a  polygon  of  12  sides,  whose  perimeters  are  equal  to 
each  other,  and  whose  sides  are  to  each  other  in  the  ratio 
of  two.     And  so  on  to  infinity. 


12 


Important  Mathematical  Discoveries 


PROPOSITION  B. 


In  a  square,  an  equilateral  triangle  and  in  all  regular 
polygons,  the  sum  of  the  least  and  greatest  radii,  divided 
by  two,  is  equal  to  the  least  radius  of  a  polygon  of  twice 
the  number  of  sides  and  equal  perimeter. 


Figure  E. 


In  Figure  E,  let  the  square  BCDE,  and  the  octagon 
FHIJ  have  equal  perimeters  (Proposition  A),  and  we  see 
that  the  line  AP  is  the  radius  of  the  inscribed  circle,  and 
AC  is  the  radius  of  the  circumscribed  circle,  and  they  are 
the  least  and   greatest  lines   that  can   be  drawn   from   the 


Preliminary  Demonstrations 


13 


center  A  to  the  perimeter  of  the  square  BCDE,  and  for 
convenience  we  will  call  them  the  least  and  greatest  radii 
ot    that  square. 

Now  the  line  NI  bisects  the  line  AP  at  d,  and  AC  at  L. 
Hence  dL  is  parallel  to  PC,  and  AL  =  LC,  and  LC  =    LH; 


Figure  F. 


therefore  AL  =  LH,  and  as  LH  =  dn,  hence  AL  =  dn. 
Then  (AP  +  AC)  h-  2  =  Ad  +  AL  =  An,  which  is  the 
least  radius  of  the  octagon  FHIJ. 

Let  Figure  F  represent  a  polygon  of  16  sides  drawn  upon 
the  square  and  octagon,  their  perimeters  being  equal.    Then 


14  Important  Mathematical  Discoveries 

in  the  octagon,  An  and  AH  are  the  least  and  greatest  radii, 
and  it  will  be  seen  that  (An  +  AH)  ^2  =  At;  and  At  is 
the  least  radius  of  a  polygon  of  16  sides,  whose  perimeter 
is  equal  to  that  of  the  octagon,  or  of  the  square,  and  so  on 
until  the  sides  become  infinite  in  number  and  minuteness. 


PROPOSITION   C. 

If  a  series  of  polygons  be  drawn  upon  each  other,  having 
the  same  center  and  equal  perimeters,  and  the  number  of 
their  sides  being  to  each  other  consecutively  in  the  ratio 
of  two,  the  greatest  radii  of  each  polygon  bisect  the  angles 
formed  at  the  center,  by  the  least  and  greatest  radii  of  its 
next  preceding  polygon,  and  the  lines  joining  the  outer 
points  of  the  greatest  radii  of  any  one  of  these  polygons, 
with  the  outer  points  of  the  greatest  radii  of  its  next  suc- 
ceeding polygon,  are  perpendicular  to  the  latter  radii. 

In  Figure  J,  let  the  square  BCDE  and  the  octagon  FHIJ, 
have  equal  perimeters,  and  let  the  line  be  be  a  side  of  a 
polygon  of  16  sides,  and  or,  a  side  of  a  polygon  of  32  sides, 
whose  perimeters  are  equal  to  that  of  the  square,  or  octa- 
gon, and  produce  the  line  An  to  X,  making  AX  equal  to 
AB  or  AC,  and  draw  the  arc  BXC,  and  it  will  be  seen  that 
the  lines  nH,  tc  and  dr,  are  each  half  a  side  of  a  polygon 
of  8,  16,  and  32  sides  respectively;  and  join  the  points  CH, 
He,   and   cr. 

■  Now  the  triangles  AnH  and  ATH  are  similar,  hence  the 
angles  nAH  and  TAH  are  equal.  And  the  line  AH  bisects 
the  angle  nAC,  and  in  like  manner  the  line  AC  bisects  the 
angle  nAH,  and  the  line  Ar  bisects  the  angle  tAc,  and  so 
on  to  infinity.  And  in  the  triangle  AHC,  the  line  Ac  is  bi- 
sected at  L,  and  LH,  LC  and  LA  are  equal  to  each  other. 
Hence  if  a  circle  be  described  on  the  line  AC  as  diameter, 
its  circumference  must  pass  through  the  point  H.  There- 
fore the  angle  AHC  is  a  right  angle. 

In  like  manner  the  angles  AcH,  Arc,  etc.,  are  right  an- 


Preliminary  Demonstrations 


15 


gles,  and  the  lines  CH,  He,  and  er  are  perpendicular  to  the 
bisecting  lines  AH,  Ac,  and  Ar  respectively.  Hence  if 
the  angles  be  bisected  indefinitely,  and  polygons  formed, 
until  the  last  polygon  that  can  possibly  be  drawn  and  re- 


Figure  J. 

tain  distinct  sides  is  reached,  the  lines  joining  the  outer 
points  of  the  greatest  radii  of  any  one  of  these  polygons, 
with  the  outer  points  of  the  greatest  radii  of  its  next  suc- 
ceeding polygon,  are  perpendicular  to  the  latter  radii,  and 
the  rectangle  formed  by  the  greatest  radii  of  any  one  of 
these  polygons  with   the  least  rad.ii  of  its  next  succeeding 


16  Important  Mathematical  Discoveries 

polygon,   is  equal   to   the   square  of  the  greatest  radius  of 
the  latter  polygon. 

It  will  be  seen  from  the  foregoing  demonstrations  that 
if  an  infinite  series  of  polygons  be  thus  formed,  the  perim- 
eter of  any  one  of  them  intersects  each  side  of  all  its  preced- 
ing polygons,  in  two  places,  the  points  of  intersection  being 
equally  distant  from  the  middle  of  the  side,  no  matter 
how  minute  the  sides  may  become,  and  as  a  circle  is  the 
ultimate  form  of  a  polygon  whose  sides  by  the  process  of 
bisection  become  infinitely  minute,  the  circumference  of 
that  circle  will  also  intersect  each  side  of  all  preceding 
polygons  in   like   manner. 


PART  11. 

PERIMETERS  AND  CIRCUMFERENCES 
MADE  EQUAL 

AND 

CURVED  LINES  MADE  EQUAL  TO 
STRAIGHT  LINES. 


PERIMETERS  AND  CIRCUMFERENCES. 
PROPOSITION  I. 

To  form  a  circle  whose  circumference  is  equal  to  the 
perimeter  of  a  given  square. 

Let  BCDE  be  a  given  square  (the  quadrature  of  the 
circle)  and  AR  its  least  radius,  and  AC  its  greatest  radius, 

FIJK  L 

H 


X 
Figure  I. 

as  in   Figure   I.     And  with  A  as  center  and  AC  as  radius, 
draw  the  arc  BFC,  and  produce  the  line  AR  to  F,  bisect- 

19 


20  Important  Mathematical  Discoveries 

ing  the  arc  BFC  at  F.  Then  bisect  the  arc  PC  at  H,  and 
FH  at  L,  and  FL  at  K,  and  FK  at  J,  and  FJ  at  I,  and  so 
on,  bisecting  to  infinity,  or  as  far  as  it  is  within  our  means 
to  bisect.     And  draw  the  lines  AH,  AL,  AK,  AJ,  and  AI. 

Then  from  point  C  draw  Ct  perpendicular  to  AH,  and 
from  the  point  t  draw  tV  perpendicular  to  AL,  and  draw 
VP,  Po,  and  on,  perpendicular  to  AK,  AJ,  and  AI,  respect- 
ively, the  perpendicular  always  falling  on  each  bisecting 
line,  until  the  least  possible  bisecting  line  is  reached,  which 
is  represented  by  AI.  Then  draw  the  final  perpendicular 
nm  to  the  line  AF,  and  it  will  be  seen  that  nm  is  half  the 
side  of  a  polygon,  whose  perimeter  is  equal  to  that 
of  the  square  BCDE.  (Prop.  C,  Preliminary  Demonstra- 
tions.) Then  through  the  point  m,  with  radius  Am,  draw 
the  circle  SmWX,  and  the  circumference  of  that  circle  is 
equal  to  the  perimeter  of  the  square  BCDE. 

Now  let  the  perpendicular  nm  be  the  least  possible  part 
or  division  of  a  straight  line,  a  mere  point,  and  we  see  that 
it  is  half  the  side  of  a  polygon  whose  perimeter  is  equal  to 
that  of  the  square,  and  whose  sides  are  capable  of  only  one 
division  or  bisection,  and  as  the  circumference  of  a  circle, 
which  is  equal  to  the  perimeter  of  a  polygon,  both  having 
the  same  center,  must  intersect  each  side  of  that  polygon 
in  two  places,  the  points  of  intersection  being  equally  dis- 
tant from  the  middle  of  the  side,  therefore,  the  circum- 
ference SmWX  must  intersect  the  perpendicular  nm,  but 
nm  being  a  mere  point,  cannot  be  either  bisected,  inter- 
sected, or  divided  further.  Therefore  the  circumference 
SmWX  must  pass  over  and  coincide  with  the  perpendic- 
ular or  point  nm,  which  represents  half  a  side  of  a  polygon, 
and  hence  it  must  pass  over  and  coincide  with  the  remain- 
ing half  of  that  side,  consequently  it  must  pass  over  and 
coincide  with  each  and  every  side  comprising  the  whole 
perimeter  of  that  polygon,  and  the  point  is  reached  where 
the  final  figure  loses  its  identity  as  a  polygon,  and  assumes 
and  contains  all  the  properties  of  a  complete  circle,  as  rep- 
resented  by  the  circle  SmWX.     Therefore,   the  circumfer- 


Perimeters  and  Circumferences 


21 


ence  of  the  circle  SmWX  is  equal  to  the  perimeter  of  the 
square  BCDE,  and  the  arc  bmd  equals  the  line  BC. 

Now  if  the  line  BC  =  2,  the  perimeter  of  the  square 
equals  8.  Hence  the  circumference  of  the  circle  SmWX 
=  8.  And  as  Am,  or  the  radius  of  a  circle  whose  circum- 
ference is  8,  is  equal  to  the  ratio  of  a  circle  to  its  circum- 
scribing square,  which  is  1.2732395  +.  Therefore  Am  = 
1.2732395  -f . 

The  following  table  gives  the  radii  of  each  polygon,  from 
the  square  to  a  polygon  of  524288  sides,  the  perimeter  of 
each  being  8,  and  the  number  of  their  sides  being  to  each 
other  consecutively  in  the  ratio  of  2. 


NO.   OF  SIDES.  SHORT    RADIUS. 

4 1. 

8 1.2071067811865475  + 

16 1.25683487303146201  + 

32 1.26914627894207004  + 

64 1.2722167067075638  + 

128 1.2729838511604253  + 

256 1.2731756083078977  + 

512 1.2732235458896797  + 

1024 1.27323553034872029  + 

2048 1.2732385264073428  + 

4096 1.2732392754215578  + 

8192 1.2732394626770475  + 

16384 1.27323950948983831  + 

32768 1.27323952119303589  + 

65536 1.2732395233334371  + 

131072 1.273239524045252  + 

262144 1.2732395242281144  + 

524288 1.27323952427383  + 

262144 1.2732395242281144  + 

524288 1.27323952427383  + 


LONG  RADIUS. 

1.414213562373095  + 
1.3065629648763765  + 
1.28145768485268807  + 
1.27528713444730576  + 
1.2737509956132868  + 
1.2733673656153701  + 
1.2732714833914617  + 
1.2732475148077608  + 
1.2732415224659654  + 
1.2732400244357728  + 
1.2732396499325372  + 
1.27323955630262909  + 
1.2732395328962334  + 
1.2732395254738383  + 
1.2722395247570668  + 
1.2732395244109769  + 
1.2732395243195456  + 

1.2732395243195456  + 


From  the  above  we  find  the  ratio  between  diameter  and 
circumference  is  3.1415927  -f. 


22 


Important  Mathematical  Discoveries 


PROPOSITION  II. 


To  form  a  circle  whose  circumference  is  equal  to  the 
perimeter  of  a  given  equilateral  triangle. 

In  Figure  II,  let  BCD  be  a  given  equilateral  triangle, 
and  draw  the  perpendicular  CE,   and  let  A  be  the  center 


^ 


Figure  II. 


of  that  triangle.  Then  AC  is  its  greatest  radius.  Then 
draw  AF  perpendicular  to  BC,  and  the  line  AF  is  its  least 
radius.     Then  produce  AF  to  L,  making  AL  =  AC.     Then 


Perimeters  and  Circumferences  23 

with  A  as  center  draw  the  arc  LC,  and  draw  the  bisecting 
lines  AH,  AI,  AJ,  AK,  etc.,  to  infinity,  or  as  far  as  it  is  pos- 
sible to  bisect.  Then  from  the  point  C  draw  Cr  perpendic- 
ular to  AH,  and  rP  perpendicular  to  AI,  and  po  to  AJ,  and 
on  to  AK,  etc.,  and  draw  the  final  perpendicular  to  the 
point  m,  on  the  line  AL.  And  with  A  as  center  and  Am 
as  radius  draw  the  circle  mS,  and  the  circunference  of  that 
circle  is  equal  to  the  perimeter  of  the  triangle  BCD.  (See 
Demonstration   of   Proposition   I,   Quadrature   of   Circle.) 


PROPOSITION  III. 

To  determine  the  Quadrature  of  the  Circle  independ- 
ently of  all  infinite  or  unlimited  bisections. 

In  Figure  III,  let  BCDE  be  a  given  square  and  A  its 
center,  and  draw  the  lines  AF  and  AD,  and  draw  DH  form- 
ing a  right  angle  with  AD,  and  produce  AF  to  H.  Then 
on  the  line  AH,  as  diameter,  draw  the  circle  HDAC.  And 
bisect  the  line  HD  at  P,  and  draw  FJ,  which  intersects 
HD  at  P,  and  draw  PA.  And  on  the  line  AH,  make  AO 
=  AP,  and  draw  PL  at  right  angles  with  AH.  Then  on 
the  line  AO  draw  the  semicircle  OnA,  and  draw  the  lines 
On  and  nA.  Then  from  the  point  D  draw  the  line  Dm 
perpendicular  to  AP,  and  draw  mR  parallel  to  FD.  And 
on  the  line  On  make  nX  =  FR,  and  draw  AX.  Then  with 
A  as  center  and  AX  as  radius,  draw  the  circle  XWYS,  and 
the  circumference  of  that  circle  is  equal  to  the  perimeter 
of  the  square  BCDE. 

Now  let  the  Hne  CD  =  2.  Then  FD  =  1.  And  AF  = 
1,  and  FH  =  1,  and  AD  and  DH  are  equal  to  each 
other,  and  each  is  equal  to  i/^  2  ,  and  Dp  =  half  the 
i/y  =  .70710678+  and  AD^  +JDP  =  AP-  =  2.5.  Hence 
A02  =  2.5.  Therefore  AO  =  i/^2l  =  1.58113883  +  .  Now 
AO  X  AF  =  An2.  But  AO  X  AF  =  AO  X  1  =  AO.  There- 
fore, An-  =  AO  =  1.581 13883 +  .  Now  in  the  right- 
angled  triangle  ADP,  PA  X  Am  =  AD^  =  2.     Hence  PA 


0 


24 


Important  Mathematical  Discoveries 


X  Am^  =  4.  And  as  PA-  =  2.5,  then  4  --  2.5  =  Am^. 
Hence  Am^  =  1.6,  and  AR^  =  1.44,  and  AR  =  1.2.  Then 
FR  =  .2,  and  hence  nX  =  .2.     Now  An^  =  1.58113883  +  , 

H 


y 


Figure  III. 


and  nX2  =  .04.  Hence  AX^  =  1.58113883  +  ,  +  .04  = 
1.62113883  +  ,  and  Ax  =  i/-"  1.62 11 3883+  =  1. 2732395  +  , 
which  is  the  ratio  of  a  circle  to  its  circumscribing  square. 


Perimeters  and  Circumferences  25 

and  is  the  radius  of  tlie  circle  XWYS.  Hence  the  circum- 
ference of  that  circle  is  equal  to  the  perimeter  of  the  square 
BCDE.  And  4  --  1.2732395+  =  3.141592  +  ,  which  is  the 
ratio  of  diameter  to  circumference. 

From  the  foregoing  demonstration  (Figure  III)  the  fol- 
lowing rule  is  obtained,  viz.: 

Multiply  the  square  of  the  radius  of  the  inscribed  circle 
by  the  i^  2^ ,  and  add  to  the  product  the  one-hundredth 
part  of  the  area  of  the  square,  and  the  square  root  of  the 
sum  is  the  radius  of  the  circle  whose  circumference  is  equal 
to  the  perimeter  of  the  given  square. 

It  will  also  be  seen  that  to  draw  a  circle  whose  circum- 
ference is  equal  to  the  perimeter  of  a  given  polygon,  bring 
that  perimeter  into  the  form  of  a  square,  and  proceed  as 
in  Figure  III  or  Figure  I,  Part  II. 


26 


Important  Mathematical  Discoveries 


PROPOSITION  IV. 

To  find  the  number  of  degrees  in  any  angle  of  a  given 
triangle. 

In  Figure  IV,  let  the  angle  BAC  be  a  given  angle,  and  it 
is  required  to  determine  the  number  of  degrees  in  it.  Draw 
the  lines  AB  and  AC  equal  to  each  other  and  draw  BC, 
and  draw  the  arc  niHn  equal  to  the  line  BC.     Then  bisect 

R 


C 


angle  BAC  by  the  line  AH,  which  is  the  radius  of  the  cir- 
cle of  which  the  arc  mHn  is  a  part.  Then  find  the  circum- 
ference of  that  circle,  and  divide  by  the  arc  mHn  or  the 
line  BC,  and  then  divide  the  result  into  360  for  the  degrees 
in  said  angle. 


PART  III. 

THE  TRISECTION  OF  AN 
ANGLE. 


THE   TRISECTION   OF   AN    ANGLE 

PROPOSITION  I. 

In  any  triangle,  the  sum  of  any  two  of  its  sides,  is  to  the 
third  side  as  either  one  of  those  two  sides  is  to  that  corres- 
ponding part  of  the  third,  cut  off  by  a  line  bisecting  the 
angle  which  the  said  two  sides  contain. 


o 

Figure  A. 

Let  BAC  (Figure  A)  be  a  triangle,  and  produce  BA  to 
E,  making  AE  =  AC,  and  draw  EC,  and  draw  AP  par- 
allel to  BC.  Then  draw  AD  parallel  to  EC,  and  AD  bi- 
sects the  angle  BAC,  as  will  be  seen  by  the  following  dem- 
onstration : 


29 


30 


Important  Mathematical  Discoveries 


AE  =  AC,  then  BE  =  BA  +  AC,  and  the  angle  AEC  = 
ACE,  and  as  the  angle  BAC  =  angles  AEC  +  ACE,  there- 
fore the  angle  BAC  =  2  ACE,  and  as  AD  is  parallel  to 
EC,    the    angles    DAC    and    ACE    are    equal.     Hence    the 


Figure  B. 


the  angle  BAC  =  2  DAC,  and  the  line  AD  bisects  the 
angle  BAC.  Now  BE  =  BA  +  AC,  and  in  triangles  BEC 
and  BAD,  BE  :  BC  :  :  BA  :  BD,  and  BE  :  BC  :  :  AE  :  AP. 
Hence  BE  :  BC  :  :  AE  :  DC.  Therefore,  BA  +  AC  :  BC 
:  :  BA  :  BD,  and  BA  +  AC  :  BC  :  :  AC  :  CD. 


The  Trisection  of  an  Angle 


31 


PROPOSITION   II.* 

In  Figure  C,  let  ABC  be  a  right-angled  isoceles  triangle 
and  it  is  required  to  trisect  the  right  angle  ABC.  Bisect 
the  angle  ABC  by  the  line  BD,  and  bisect  and  angle  BAC 
by  the  line  AP,  and  draw  BH  =  BP,  and  with  A  as  center 
and  AH  as  radius,  draw  the  arc  HS,  meeting  the  line  DB 


J 


yn 


D  ^ 

Figure  C. 


K 


C 


produced,  at  the  point  S,  and  draw  AS  and  CS.  Then 
draw  the  line  SJ  bisecting  the  angle  DSA,  and  draw  SK 
bisecting  the  angle  DSC,  and  draw  Bm  and  Bn  parallel 
to  SJ  and  SK  respectively,  and  the  lines  Bm  and  Bn  tri- 
sect the  angle  ABC. 


*The  trisection  of  acute  and  obtuse  angles  will  appear  in  the   next 
issue  of  book. 


32  Important   Mathematical  Discoveries 

Now  in  triangle  ABC,  let  the  line  Ac  =  2.  Hence  DB, 
DA  and  DC  each  equals  1,  and  AB  and  CB  each  equals 
i/T.  And  as  the  line  AP  bisects  the  angle  BAC,  it  will  be 
seen  that  the  line  BP  =  .58578  +  ,  and  as  BP  =  BH,  there- 
fore the  line  AH  =  (i/T+  .585^8  +  )  =  2.  Hence  the  line 
AS  =  2,  and  the  line  SD  =  i/  3  ,  and  the  triangle  ASC  is 
equilateral,  and  the  angle  ASC  is  60  degrees,  the  angle 
ASD  is  30  degrees,  and  the  angle  JSD  =  15  degrees.  And 
as  the  dine  Bm  is  parallel  to  SJ,  hence  the  angle  mBD  is 
15  degrees.  And  as  the  angle  ABD  =  45  degrees,  there- 
fore the  angle  niBD  =  one-third  of  angle  ABD.  In  like 
manner  the  angle  nBD  =  one-third  of  angle  CBD.  Hence 
the  angle  niBN  is  one-third  of  the  angle  ABC. 

In  all  cases  an  isoceles  triangle  must  be  formed,  the  angle 
to  be  trisected  being  the  vertical  angle,  as  in  Figure  C. 


PART  IV. 

THE  QUADRATURE  AND  DUPLICATION 
OF  THE  CUBE. 


QUADRATURE  AND  DUPLICATION 
OF  THE  CUBE. 

PROPOSITION  I. 

In  Figure  I,  let  ABCD  be  a  square  side  of  a  given  cube, 
and  on  the  line  AB  draw  the  semicircle  BOA,  and  let  BE 
=  1,  and  draw  the  lines  EFP,  FB  and  FA.     Then  AB  X 


T 


5 


D 

P 

/                   / 
/                X 
1               X 
1          X 

/ 

/ 
/ 

/ 

"H 

vA 

R 


c 


A 
Figure  I, 


L 


BE  =  FB-  =  AB.  Then  make  F^  =  BF,  and  draw  LJ. 
Then  AB  X  BL  =  BJ^  =  BH^,  and  draw  HS  at  right  angles 
with    BH,    and    draw    the    square    SBRT,    and    the    square 

.^5 


36 


Important  Mathematical  Discoveries 


SBRT  =  AB^   of  which   the  square  ABCD   is  one  of  the 
square  sides. 


F    P  D 


Figure  II. 


Now   BE  =  1,   then   AB  X  BE^BF^  =  AB.     And    BF 

=  j/AB,    and    therefore    BL  =  i/AB.     Then   AB  X  BL  - 

i/AB^.     But    AB  X  BL  =  BJ-'  =  BH^  =  BS.       Hence    BS 

=  i/AB'  ,     consequently    BS^  =  AB^.     Hence   the    square 

SBRT  =  AB\ 


Quadrature  and  Duplication  of  the  Cube         37 


PROPOSITION  II. 


To  bring  a  slab  one  inch  thick  into  the  form  of  a  cube, 
whose    soHd    contents   are   equal    to    the    solid    contents   of 

said  cube. 

In  Figure  II,  let  ABYX  be  a  square  slab  one  inch  thick, 
whose  solid   contents  =  27.     Then  the    line    AB  =  i.^27  = 


H  F  ^ 

Figure  III. 


^    E 


5.196  +  .  Now  let  BD  =  1,  then  AB  X  BD  =  BC^  (the 
semicircle  BSA  being  drawn).  Then  make  BE  =  BC,  and 
draw  the  lines  EL,  LB  and  CA.     Then  make  BP  =  i/  BE, 


38 


Important  Mathematical  Discoveries 


and  it  is  also  the  4th  root  of  AB.  Then  AB  X  BP  =  BR^. 
And  make  the  angle  LBS  =  two-thirds  of  the  angle  LBR, 
and  draw  SF,  then  the  line  BS  is  the  cube  root  of  the  slab 
ABYX,   and  is  therefore  the  cube  root  of  27,   which  is  3. 


A 


^  H        E      F  P     D 

Figure  IV. 


Then  make  BW  =  BS,  and  on  BW  draw  the  semi-cir- 
cle BmW,  and  draw  the  lines  Bm  and  niW.  Now  AB  X 
BF  =  BS2  =  BW2  =  9.  Hence  9  ^  5.196+  =  i  3",+ 
then  BF  =  1.732+  and  =  i/BW.  Hence  BW  is  the  re- 
ciuired  line  representing  the  edge  of  the  cube  which  is  equal 
to  the  given  slab;  and  BF^  =  AB.  Hence  any  rectangu- 
lar slab  may  be  brought  into  a  cube. 


Quadrature  and  Duplication  of  the  Cube         39 

Or,  let  Figure  III  be  a  reproduction  of  Figure  II,  then 
BA  multiplied  by  its  square  root  AF  =  AL-.  Then  make 
An  =  the  8th  root  of  BA.  Then  BA  X  An  =  AP.  And 
the  difference  between  the  angles  thus  formed  at  A,  which 
is  the  angle  LAI,  bisect  by  line  AR,  and  draw  the  line  Rm. 
Then  Am  is  the  cube  root  of  AB. 

This  method  is  different  from  that  of  Figure  II,  and  has 
the  advantage  of  dispensing  with  trisections. 


PROPOSITION  III. 

To  form  a  cube  whose  solid  contents  are  twice  the  solid 
contents  of  a  given  cube,  and  vice  versa. 

In  Figure  IV,  let  the  square  BHVK  be  a  side  of  a  given 
cube,  and  let  BJTO  represent  a  square  slab  one  inch  thick, 
and  equal  to  the  contents  of  the  given  cube.  Then  pro- 
duce BJ  to  A,  making  BA  =  to  the  diagonal  of  the  square 
BJTO,  and  form  the  square  ABYX,  and  it  will  be  seen 
that  the  square  ABYX  is  twice  the  square  BJTO.  Then 
bring  the  square  ABYX  into  a  cube  as  represented  by  the 
square  side  BHtb. 

Now  the  square  ABYX  is  twice  the  square  BJTO.  It  is 
therefore  twice  the  cube  BHVK,  and  the  square  ABYX 
being  equal  to  the  cube  represented  by  the  square  side 
BHtb,  hence  the  cube  BHtb  is  twice  the  cube  BHVK. 

It  will  be  seen  that  a  cube  may  be  formed  whose  solid 
contents  are  any  given  number  of  times  greater  than  a  given 
cube. 


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